An-Najah N. University Faculty of Engineering and Information Technology Department of Management Information systems Operations Research and Applications MIS: Prepared by Mr.Maher Abubaker Fall 2015/2016 Resources Daniel J. Epstein Department of Industrial and Systems Engineering University of Southern California Andrew and Erna Viterbi School of Engineering Operations Research: An Introduction, 9/EHamdy A. Taha, University of ArkansasISBN-10: X ISBN-13: ©2011 Prentice Hall Cloth, 832 ppPublished 08/29/ Introduction/ page#downlaoddiv ™ INFORMS – ™ ORMS - ™ Science of Better -

Duality To every LP, there is a dual! A decision without optimization is like an arch without a keystone. …olde English saying circa 1871

Primal problem Max C t X AX <= b X>=0 Dual problem Min b t Y AY >= c Y>=0 From Primal To Dual Problem

A Primal Problem Let X = number of units of product A to manufacture x2 = number of units of product B to manufacture Max Profit = z = 5 x x 2 subject to: x x 2 <= 20 (labor-hours) 2 x x 2 <= 30 (machine hours) 6 x x 2 <= 180 ($ - raw material) x 1 >= 0, x 2 >= 0

The Dual Let y 1 = implicit value (cost) of one labor-hour y 2 = implicit value (cost) of one machine hour y 3 = implicit value (cost) of $1.00 for raw material Min y 0 = 20y y y 3 subject to: y 1 + 2y y 3 >= 5 2y 1 + 2y y 3 >= 15 y 1, y 2, y 3 >=0 Max z = 5 x x 2 subject to: x x 2 <= 20 2 x x 2 <= 30 6 x x 2 <= 180 From max to min From less than to greater than

It is quite clear now. The dual is assigning a (implicit) cost to each resource that minimizes the total cost of the available resources while insuring that the cost of the resources required to produce one unit of a product is at least as great as the profit obtained from that product. Gosh. It really doesn’t get any better than this does it? Min y 0 = 20y y y 3 subject to: y 1 + 2y y 3 >= 5 2y 1 + 2y y 3 >= 15 y 1, y 2, y 3 >=0

Aren’t there other names for the values of the dual variables? Dual Variables Implicit Values Implicit Costs Shadow Prices Opportunity Costs

Max z = 6x 1 + 4x 2 Subj. to: x 1 + x 2 <= 12 x 1 -2x 2 <= 6 x 2 <= 8 x 1, x 2 >= 0 Primal Min y 0 = 12y 1 + 6y 2 + 8y 3 subject to: y 1 + y 2 >= 6 y 1 - 2y 2 + y 3 >= 4 y 1, y 2, y 3 >= 0 Dual

The Final Tableau to the Primal Basic Eq Varzx 1 x 2 s1s2s3 RHS 0z 1x 2 2x 1 3s3 0102/31/ /31/ /3-1/ /32/3068 y1y1 y2y2 y3y3 y0y0

Initial & Final Tableau for Dual Problem Basic Eq Vary 0 y 1 y 2 y 3 y 4 y 5 y 6 RHS 0z M0 1x y Basic Eq Vary 0 y 1 y 2 y 3 y 4 y 5 y 6 RHS 0z y /3-1/31/3 -2/3 2y /3-2/3-1/3-16/3

Primal and Dual Problems